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Hot object in a vacuum
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$begingroup$
A friend of mine told me that if you were to stand beside plate of metal that is millions of degrees hot, inside a 100% vacuum, you would not feel its heat. Is this true? I understand the reasoning that there is no air, thus no convection, and unless you're touching it, there's no conduction either. I'm more so asking about thermal radiation emitted by it.
thermodynamics temperature thermal-radiation vacuum
edited 7 hours ago
Qmechanic♦
111k12 gold badges214 silver badges1307 bronze badges
asked 9 hours ago
Peter_BrowningPeter_Browning
711 silver badge3 bronze badges
$endgroup$
add a comment |
$begingroup$
A friend of mine told me that if you were to stand beside plate of metal that is millions of degrees hot, inside a 100% vacuum, you would not feel its heat. Is this true? I understand the reasoning that there is no air, thus no convection, and unless you're touching it, there's no conduction either. I'm more so asking about thermal radiation emitted by it.
thermodynamics temperature thermal-radiation vacuum
edited 7 hours ago
Qmechanic♦
111k12 gold badges214 silver badges1307 bronze badges
asked 9 hours ago
Peter_BrowningPeter_Browning
711 silver badge3 bronze badges
$endgroup$
6
$begingroup$
Your friend is wrong. Let him stand there and see what happens. He will be fried.
$endgroup$
– Chet Miller
9 hours ago
$begingroup$
That's a weird thing for your friend to argue against. High temperature differences are when you expect heat transfer due to radiation to be the dominant factor.
$endgroup$
– JMac
8 hours ago
add a comment |
$begingroup$
A friend of mine told me that if you were to stand beside plate of metal that is millions of degrees hot, inside a 100% vacuum, you would not feel its heat. Is this true? I understand the reasoning that there is no air, thus no convection, and unless you're touching it, there's no conduction either. I'm more so asking about thermal radiation emitted by it.
thermodynamics temperature thermal-radiation vacuum
edited 7 hours ago
Qmechanic♦
111k12 gold badges214 silver badges1307 bronze badges
asked 9 hours ago
Peter_BrowningPeter_Browning
711 silver badge3 bronze badges
$endgroup$
A friend of mine told me that if you were to stand beside plate of metal that is millions of degrees hot, inside a 100% vacuum, you would not feel its heat. Is this true? I understand the reasoning that there is no air, thus no convection, and unless you're touching it, there's no conduction either. I'm more so asking about thermal radiation emitted by it.
thermodynamics temperature thermal-radiation vacuum
thermodynamics temperature thermal-radiation vacuum
edited 7 hours ago
Qmechanic♦
111k12 gold badges214 silver badges1307 bronze badges
asked 9 hours ago
Peter_BrowningPeter_Browning
711 silver badge3 bronze badges
edited 7 hours ago
Qmechanic♦
111k12 gold badges214 silver badges1307 bronze badges
asked 9 hours ago
Peter_BrowningPeter_Browning
711 silver badge3 bronze badges
edited 7 hours ago
Qmechanic♦
111k12 gold badges214 silver badges1307 bronze badges
edited 7 hours ago
Qmechanic♦
111k12 gold badges214 silver badges1307 bronze badges
edited 7 hours ago
Qmechanic♦
111k12 gold badges214 silver badges1307 bronze badges
111k12 gold badges214 silver badges1307 bronze badges
asked 9 hours ago
Peter_BrowningPeter_Browning
711 silver badge3 bronze badges
asked 9 hours ago
Peter_BrowningPeter_Browning
711 silver badge3 bronze badges
asked 9 hours ago
Peter_BrowningPeter_Browning
711 silver badge3 bronze badges
711 silver badge3 bronze badges
6
$begingroup$
Your friend is wrong. Let him stand there and see what happens. He will be fried.
$endgroup$
– Chet Miller
9 hours ago
$begingroup$
That's a weird thing for your friend to argue against. High temperature differences are when you expect heat transfer due to radiation to be the dominant factor.
$endgroup$
– JMac
8 hours ago
add a comment |
6
$begingroup$
Your friend is wrong. Let him stand there and see what happens. He will be fried.
$endgroup$
– Chet Miller
9 hours ago
$begingroup$
That's a weird thing for your friend to argue against. High temperature differences are when you expect heat transfer due to radiation to be the dominant factor.
$endgroup$
– JMac
8 hours ago
6
6
$begingroup$
Your friend is wrong. Let him stand there and see what happens. He will be fried.
$endgroup$
– Chet Miller
9 hours ago
$begingroup$
Your friend is wrong. Let him stand there and see what happens. He will be fried.
$endgroup$
– Chet Miller
9 hours ago
$begingroup$
That's a weird thing for your friend to argue against. High temperature differences are when you expect heat transfer due to radiation to be the dominant factor.
$endgroup$
– JMac
8 hours ago
$begingroup$
That's a weird thing for your friend to argue against. High temperature differences are when you expect heat transfer due to radiation to be the dominant factor.
$endgroup$
– JMac
8 hours ago
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
I'm more so asking about thermal radiation emitted by it.
Here's a quantitative estimate.
Suppose that the hot plate remained intact long enough to do the experiment. For a rough estimate, we can treat the hot metal plate as a blackbody. According to Wien's displacement law, the electromagnetic radiation emitted by a blackbody at temprature $T$ is strongest at the wavelength
$$
lambda = frac{b}{T}
hskip1cm
bapprox 2.9times 10^{-3} text{ meter}cdottext{Kelvin}.
tag{1}
$$
The total power emitted per unit area is given by the Stefan-Boltzmann law
$$
frac{P}{A}= sigma T^4
hskip1cm
sigmaapprox 5.7times 10^{-8}
frac{text{Watts}}{text{meter}^2cdottext{Kelvin}^4}.
tag{2}
$$
For $T=10^6$ Kelvin, these estimates give
$$
lambdaapprox 2.9times 10^{-9}text{ meter}
$$
and
$$
frac{P}{A}approx 5.7times 10^{16} frac{text{Watts}}{text{meter}^2}.
$$
This wavelength is in the X-ray range, and this power level is more than a trillion times the power a person on earth would receive from the sun if there were no clouds and no air.
Would you feel it? I'm not sure. Probably only very briefly.
answered 8 hours ago
Chiral AnomalyChiral Anomaly
16.8k2 gold badges22 silver badges53 bronze badges
$endgroup$
add a comment |
$begingroup$
Your friend is completely wrong. Consider the following things:
The temperature that you are talking about is very high, no metal would be in a solid state at the temperature you are talking about. So, before your plate reaches millions of degrees, it would have melted long before.
Your understanding is correct in terms of thermal radiation. The radiation of Sun reaches Earth and there is a vacuum between. So, if you have an object as hot as you are talking about it will emit thermal radiation energy per unit time as per the Stefan-Boltzmann Equation. And remember, the rate of emitted radiation is proportional to the fourth power of temperature, so doubling the temperature would increase the rate by 16 times. You can calculate the energy reaching per unit area of your skin and find out what will happen!
answered 8 hours ago
orionorion
7762 gold badges8 silver badges15 bronze badges
$endgroup$
add a comment |
$begingroup$
You would feel its blackbody radiation as it is an EM wave and does not need a physical support to propagate itself. Also, "100% vaccuum" is not rigorous definition of the state of your system.
answered 9 hours ago
PackSciencesPackSciences
25310 bronze badges
$endgroup$
add a comment |
$begingroup$
Your friend is correct, you would not feel it's heat. But only because you would be dead in an instant.
answered 26 mins ago
JohnnyBJohnnyB
1
New contributor
JohnnyB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I'm more so asking about thermal radiation emitted by it.
Here's a quantitative estimate.
Suppose that the hot plate remained intact long enough to do the experiment. For a rough estimate, we can treat the hot metal plate as a blackbody. According to Wien's displacement law, the electromagnetic radiation emitted by a blackbody at temprature $T$ is strongest at the wavelength
$$
lambda = frac{b}{T}
hskip1cm
bapprox 2.9times 10^{-3} text{ meter}cdottext{Kelvin}.
tag{1}
$$
The total power emitted per unit area is given by the Stefan-Boltzmann law
$$
frac{P}{A}= sigma T^4
hskip1cm
sigmaapprox 5.7times 10^{-8}
frac{text{Watts}}{text{meter}^2cdottext{Kelvin}^4}.
tag{2}
$$
For $T=10^6$ Kelvin, these estimates give
$$
lambdaapprox 2.9times 10^{-9}text{ meter}
$$
and
$$
frac{P}{A}approx 5.7times 10^{16} frac{text{Watts}}{text{meter}^2}.
$$
This wavelength is in the X-ray range, and this power level is more than a trillion times the power a person on earth would receive from the sun if there were no clouds and no air.
Would you feel it? I'm not sure. Probably only very briefly.
answered 8 hours ago
Chiral AnomalyChiral Anomaly
16.8k2 gold badges22 silver badges53 bronze badges
$endgroup$
add a comment |
$begingroup$
I'm more so asking about thermal radiation emitted by it.
Here's a quantitative estimate.
Suppose that the hot plate remained intact long enough to do the experiment. For a rough estimate, we can treat the hot metal plate as a blackbody. According to Wien's displacement law, the electromagnetic radiation emitted by a blackbody at temprature $T$ is strongest at the wavelength
$$
lambda = frac{b}{T}
hskip1cm
bapprox 2.9times 10^{-3} text{ meter}cdottext{Kelvin}.
tag{1}
$$
The total power emitted per unit area is given by the Stefan-Boltzmann law
$$
frac{P}{A}= sigma T^4
hskip1cm
sigmaapprox 5.7times 10^{-8}
frac{text{Watts}}{text{meter}^2cdottext{Kelvin}^4}.
tag{2}
$$
For $T=10^6$ Kelvin, these estimates give
$$
lambdaapprox 2.9times 10^{-9}text{ meter}
$$
and
$$
frac{P}{A}approx 5.7times 10^{16} frac{text{Watts}}{text{meter}^2}.
$$
This wavelength is in the X-ray range, and this power level is more than a trillion times the power a person on earth would receive from the sun if there were no clouds and no air.
Would you feel it? I'm not sure. Probably only very briefly.
answered 8 hours ago
Chiral AnomalyChiral Anomaly
16.8k2 gold badges22 silver badges53 bronze badges
$endgroup$
add a comment |
$begingroup$
I'm more so asking about thermal radiation emitted by it.
Here's a quantitative estimate.
Suppose that the hot plate remained intact long enough to do the experiment. For a rough estimate, we can treat the hot metal plate as a blackbody. According to Wien's displacement law, the electromagnetic radiation emitted by a blackbody at temprature $T$ is strongest at the wavelength
$$
lambda = frac{b}{T}
hskip1cm
bapprox 2.9times 10^{-3} text{ meter}cdottext{Kelvin}.
tag{1}
$$
The total power emitted per unit area is given by the Stefan-Boltzmann law
$$
frac{P}{A}= sigma T^4
hskip1cm
sigmaapprox 5.7times 10^{-8}
frac{text{Watts}}{text{meter}^2cdottext{Kelvin}^4}.
tag{2}
$$
For $T=10^6$ Kelvin, these estimates give
$$
lambdaapprox 2.9times 10^{-9}text{ meter}
$$
and
$$
frac{P}{A}approx 5.7times 10^{16} frac{text{Watts}}{text{meter}^2}.
$$
This wavelength is in the X-ray range, and this power level is more than a trillion times the power a person on earth would receive from the sun if there were no clouds and no air.
Would you feel it? I'm not sure. Probably only very briefly.
answered 8 hours ago
Chiral AnomalyChiral Anomaly
16.8k2 gold badges22 silver badges53 bronze badges
$endgroup$
I'm more so asking about thermal radiation emitted by it.
Here's a quantitative estimate.
Suppose that the hot plate remained intact long enough to do the experiment. For a rough estimate, we can treat the hot metal plate as a blackbody. According to Wien's displacement law, the electromagnetic radiation emitted by a blackbody at temprature $T$ is strongest at the wavelength
$$
lambda = frac{b}{T}
hskip1cm
bapprox 2.9times 10^{-3} text{ meter}cdottext{Kelvin}.
tag{1}
$$
The total power emitted per unit area is given by the Stefan-Boltzmann law
$$
frac{P}{A}= sigma T^4
hskip1cm
sigmaapprox 5.7times 10^{-8}
frac{text{Watts}}{text{meter}^2cdottext{Kelvin}^4}.
tag{2}
$$
For $T=10^6$ Kelvin, these estimates give
$$
lambdaapprox 2.9times 10^{-9}text{ meter}
$$
and
$$
frac{P}{A}approx 5.7times 10^{16} frac{text{Watts}}{text{meter}^2}.
$$
This wavelength is in the X-ray range, and this power level is more than a trillion times the power a person on earth would receive from the sun if there were no clouds and no air.
Would you feel it? I'm not sure. Probably only very briefly.
answered 8 hours ago
Chiral AnomalyChiral Anomaly
16.8k2 gold badges22 silver badges53 bronze badges
edited 8 hours ago
answered 8 hours ago
Chiral AnomalyChiral Anomaly
16.8k2 gold badges22 silver badges53 bronze badges
answered 8 hours ago
Chiral AnomalyChiral Anomaly
16.8k2 gold badges22 silver badges53 bronze badges
answered 8 hours ago
Chiral AnomalyChiral Anomaly
16.8k2 gold badges22 silver badges53 bronze badges
16.8k2 gold badges22 silver badges53 bronze badges
add a comment |
add a comment |
$begingroup$
Your friend is completely wrong. Consider the following things:
The temperature that you are talking about is very high, no metal would be in a solid state at the temperature you are talking about. So, before your plate reaches millions of degrees, it would have melted long before.
Your understanding is correct in terms of thermal radiation. The radiation of Sun reaches Earth and there is a vacuum between. So, if you have an object as hot as you are talking about it will emit thermal radiation energy per unit time as per the Stefan-Boltzmann Equation. And remember, the rate of emitted radiation is proportional to the fourth power of temperature, so doubling the temperature would increase the rate by 16 times. You can calculate the energy reaching per unit area of your skin and find out what will happen!
answered 8 hours ago
orionorion
7762 gold badges8 silver badges15 bronze badges
$endgroup$
add a comment |
$begingroup$
Your friend is completely wrong. Consider the following things:
The temperature that you are talking about is very high, no metal would be in a solid state at the temperature you are talking about. So, before your plate reaches millions of degrees, it would have melted long before.
Your understanding is correct in terms of thermal radiation. The radiation of Sun reaches Earth and there is a vacuum between. So, if you have an object as hot as you are talking about it will emit thermal radiation energy per unit time as per the Stefan-Boltzmann Equation. And remember, the rate of emitted radiation is proportional to the fourth power of temperature, so doubling the temperature would increase the rate by 16 times. You can calculate the energy reaching per unit area of your skin and find out what will happen!
answered 8 hours ago
orionorion
7762 gold badges8 silver badges15 bronze badges
$endgroup$
add a comment |
$begingroup$
Your friend is completely wrong. Consider the following things:
The temperature that you are talking about is very high, no metal would be in a solid state at the temperature you are talking about. So, before your plate reaches millions of degrees, it would have melted long before.
Your understanding is correct in terms of thermal radiation. The radiation of Sun reaches Earth and there is a vacuum between. So, if you have an object as hot as you are talking about it will emit thermal radiation energy per unit time as per the Stefan-Boltzmann Equation. And remember, the rate of emitted radiation is proportional to the fourth power of temperature, so doubling the temperature would increase the rate by 16 times. You can calculate the energy reaching per unit area of your skin and find out what will happen!
answered 8 hours ago
orionorion
7762 gold badges8 silver badges15 bronze badges
$endgroup$
Your friend is completely wrong. Consider the following things:
The temperature that you are talking about is very high, no metal would be in a solid state at the temperature you are talking about. So, before your plate reaches millions of degrees, it would have melted long before.
Your understanding is correct in terms of thermal radiation. The radiation of Sun reaches Earth and there is a vacuum between. So, if you have an object as hot as you are talking about it will emit thermal radiation energy per unit time as per the Stefan-Boltzmann Equation. And remember, the rate of emitted radiation is proportional to the fourth power of temperature, so doubling the temperature would increase the rate by 16 times. You can calculate the energy reaching per unit area of your skin and find out what will happen!
answered 8 hours ago
orionorion
7762 gold badges8 silver badges15 bronze badges
answered 8 hours ago
orionorion
7762 gold badges8 silver badges15 bronze badges
answered 8 hours ago
orionorion
7762 gold badges8 silver badges15 bronze badges
answered 8 hours ago
orionorion
7762 gold badges8 silver badges15 bronze badges
7762 gold badges8 silver badges15 bronze badges
add a comment |
add a comment |
$begingroup$
You would feel its blackbody radiation as it is an EM wave and does not need a physical support to propagate itself. Also, "100% vaccuum" is not rigorous definition of the state of your system.
answered 9 hours ago
PackSciencesPackSciences
25310 bronze badges
$endgroup$
add a comment |
$begingroup$
You would feel its blackbody radiation as it is an EM wave and does not need a physical support to propagate itself. Also, "100% vaccuum" is not rigorous definition of the state of your system.
answered 9 hours ago
PackSciencesPackSciences
25310 bronze badges
$endgroup$
add a comment |
$begingroup$
You would feel its blackbody radiation as it is an EM wave and does not need a physical support to propagate itself. Also, "100% vaccuum" is not rigorous definition of the state of your system.
answered 9 hours ago
PackSciencesPackSciences
25310 bronze badges
$endgroup$
You would feel its blackbody radiation as it is an EM wave and does not need a physical support to propagate itself. Also, "100% vaccuum" is not rigorous definition of the state of your system.
answered 9 hours ago
PackSciencesPackSciences
25310 bronze badges
answered 9 hours ago
PackSciencesPackSciences
25310 bronze badges
answered 9 hours ago
PackSciencesPackSciences
25310 bronze badges
answered 9 hours ago
PackSciencesPackSciences
25310 bronze badges
25310 bronze badges
add a comment |
add a comment |
$begingroup$
Your friend is correct, you would not feel it's heat. But only because you would be dead in an instant.
answered 26 mins ago
JohnnyBJohnnyB
1
New contributor
JohnnyB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Your friend is correct, you would not feel it's heat. But only because you would be dead in an instant.
answered 26 mins ago
JohnnyBJohnnyB
1
New contributor
JohnnyB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Your friend is correct, you would not feel it's heat. But only because you would be dead in an instant.
answered 26 mins ago
JohnnyBJohnnyB
1
New contributor
JohnnyB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Your friend is correct, you would not feel it's heat. But only because you would be dead in an instant.
answered 26 mins ago
JohnnyBJohnnyB
1
New contributor
JohnnyB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 26 mins ago
JohnnyBJohnnyB
1
New contributor
JohnnyB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 26 mins ago
JohnnyBJohnnyB
1
answered 26 mins ago
JohnnyBJohnnyB
1
1
New contributor
JohnnyB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
JohnnyB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
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6
$begingroup$
Your friend is wrong. Let him stand there and see what happens. He will be fried.
$endgroup$
– Chet Miller
9 hours ago
$begingroup$
That's a weird thing for your friend to argue against. High temperature differences are when you expect heat transfer due to radiation to be the dominant factor.
$endgroup$
– JMac
8 hours ago